Question

A square ABCD of side 1 mm is kept at distance 15 cm in front of the concave mirror as shown in figure. The focal length of the mirror is 10 cm. The length of the perimeter of its image will be

• A. 8 mm
• B. 2 mm
• C. 12 mm
• D. 6 mm

Answer 1

The correct option is C 12 mm

Here $AB$ and $CD$ will undergo longitudinal magnification and $AD$ and $BC$ will undergo lateral magnification.

For $CD$,

Magnification $m=\frac{f}{f-u}=\frac{-10}{15-10}=-2$
Therefore $C^{\prime }{D}^{\prime }=2\times 1=2mm$

Similarly, $A^{\prime }{B}^{\prime }=2\times 1=2mm$

For $BC$,

$v_i=-{\left(\frac{v}{u}\right)}^2{v}_0$

$\frac{\Delta v}{\Delta t}=-{\left(\frac{v}{u}\right)}^2\frac{\Delta u}{\Delta t}$

$\frac{\Delta v}{\Delta u}={(-\frac{v}{u})}^2=\frac{f^2}{(f-u{)}^2}$

$\because [-\frac{v}{u}=m=\frac{f}{f-u}]$

$\Rightarrow \frac{\Delta v}{\Delta u}=2^2=4$

Therefore for $BC,AD$,

$B^{\prime }{C}^{\prime }=A^{\prime }{D}^{\prime }=4\times 1=4mm$

Perimeter $A^{\prime }{B}^{\prime }{C}^{\prime }{D}^{\prime }=2+2+4+4=12mm$